Gradient fields in simply connected space

Question

Consider there following statement:

Let F be a vector field such that $\nabla \times \textbf{F} = 0$ over a simply connected region R. Then in R, there is a scalar potential function $\phi$ such that $\textbf{F} = \nabla \phi$.

How would you go about justifying that R must be simply connected?

Answer 1

You wouldn't, because the set $\mathbf R$ need not be simply connected.

Take any smooth function $f$ defined on the whole of $\mathbb R^2$, and consider the restriction $\mathbf F$ of the gradient of $f$ to $\mathbf R= \mathbb R^2\setminus\{0\}$, which is a non-simply-connected open set. Then $\mathbf F$ does have a potential!